leetcode:25 reverse Nodes in k-Group python

最新推荐文章于 2020-10-16 21:03:21 发布

原创最新推荐文章于 2020-10-16 21:03:21 发布 · 107 阅读

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LeetCode题解同时被 3 个专栏收录

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算法学习

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链表

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本文介绍了一种链表操作算法——K组翻转。该算法将链表按K个一组进行翻转，并提供了完整的Python实现代码。通过构造示例链表并运行翻转函数，验证了算法的有效性和正确性。

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class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

    def __str__(self):
        return 'val:%d' % self.val
class Solution:
    def reverseKGroup(self, head,  k):
        dummy = ListNode(0)
        dummy.next = head

        pre, tail = dummy, dummy
        while tail.next:
            count = k
            while count and tail:
                tail, count = tail.next, count-1

            if not tail:
                break
            #(pre->start,tail->next)
            start = pre.next
            next = tail.next
            tail.next = None
            #翻转k个节点,即[start,tail]并连接好
            pre.next = self.reverseList(start)
            #[tail,start]--->next
            start.next = next
            #归位，为下一次翻转做准备
            pre = start
            tail = pre
        return dummy.next
    def reverseList(self, head: ListNode) -> ListNode:
        cur, pre = head, None
        while cur:
            cur.next, pre, cur = pre, cur, cur.next
        return pre

if __name__ == '__main__':
    a = ListNode(1)
    b = ListNode(2)
    c = ListNode(3)
    d = ListNode(4)
    e = ListNode(5)
    f = ListNode(6)

    #构造一个简单链表
    a.next, b.next, c.next, d.next, e.next = b, c, d, e, f
    p = a
    while(p):
        print(p.val)
        p=p.next
    print("----------------")
    p=Solution().reverseKGroup(a,5)
    while(p):
        print(p.val)
        p = p.next